Solveeit Logo

Question

Chemistry Question on Enthalpy change

For complete combustion of methanol
CH3OH(I)+32O2(g)CO2(g)+2H2O(I)CH_3OH(I)+\frac 32O_2(g)→CO_2(g)+2H_2O(I)
the amount of heat produced as measured by bomb calorimeter is 726726 kJ mol–1 at 27°C27°C. The enthalpy of combustion for the reaction is x kJmol1x \ kJ mol^{–1}, where xx is ______ . (Nearest integer)
(Given: R=8.3 JK1mol1R= 8.3 \ JK^{–1}mol^{–1})

Answer

CH3OH(I)+32O2(g)CO2(g)+2H2O(I)CH_3OH(I)+\frac 32O_2(g)→CO_2(g)+2H_2O(I)
The enthalpy of combustion for the reaction,
ΔH=ΔU+ΔngRTΔH=ΔU+Δn_gRT
ΔH=726kJ+(12)×8.3×300ΔH=−726 kJ+(−\frac 12)×8.3×300
ΔH727 kJmol1ΔH≃ –727\ kJmol^{–1}

So, the answer is 727727.